Chapter 8.6, Problem 41PPSE

Solution

a.

To determine: The amount of the money spend each year on gasoline.

The required ratio is $\frac{2}{9}r$ .

Given Information:

The average distance driven is 15000 miles per year and cart gets 24 miles per gallon. The gasoline costs is $\$3.60\text{ a gallon}$ .

Calculation:

Consider the given information,

The total cost for an annual travel can be computed by multiplying the amount used to the cost per amount. In symbols,

  $T=AC$

Where T is the total cost A is the amount used, and C is the cost per amount.

The total amount used can be computed by dividing the total distance by the distance that an amount can cover.

Substitute the given values in the formula $T=\frac{DC}{d}$ , where D is the total distance d is the distance per amount, and C is the cost per amount.

  $\begin{array}{c}T=\frac{15000}{24}\cdot 3.60\\ =\$2250\end{array}$

Therefore, the total required amount is $\$2250$ .

b.

To determine: The expression to represent the new yearly cost of gasoline.

The obtained expression is $T=\frac{54000}{24+x}$ .

Given Information:

The average distance driven is 15000 miles per year and cart gets 24 miles per gallon. The gasoline costs is $\$3.60\text{ a gallon}$ .

Calculation:

Consider the given information,

The total cost for an annual travel can be computed by multiplying the amount used to the cost per amount. In symbols,

  $T=AC$

Where T is the total cost A is the amount used, and C is the cost per amount.

The total amount used can be computed by dividing the total distance by the distance that an amount can cover.

Substitute the given values in the formula $T=\frac{DC}{d}$ , where D is the total distance d is the distance per amount, and C is the cost per amount.

  $\begin{array}{c}T=\left(\frac{15000}{24+x}\right)3.6\\ =\frac{54000}{24+x}\end{array}$

Hence, the obtained result is $T=\frac{54000}{24+x}$ .

c.

To determine: The comparison to represent the total savings on gasoline per year.

The required expression is $S=\frac{2250}{24+x}$ .

Given Information:

The average distance driven is 15000 miles per year and cart gets 24 miles per gallon. The gasoline costs is $\$3.60\text{ a gallon}$ .

Calculation:

Consider the given information,

The obtained total cost in part (a) is $2250 and the part (b) is $T=\frac{54000}{24+x}$ .

To solve for the decrease in cost, subtract the total cost in part B from the total cost in part A .

  $\begin{array}{c}S=2250-\frac{54000}{24+x}\\ =\frac{54000+2250x-54000}{24+x}\\ =\frac{2250}{24+x}\end{array}$

Therefore, the obtained expression is $S=\frac{2250}{24+x}$ .

d.

To determine: The number of miles per gallon the new car get.

The required miles is $32.73$ miles par gallons.

Given Information:

The saving with new car on per year is $600.

Calculation:

Consider the given information,

Since the total savings per year is worth $600, equate the expression obtained on part c to 600.

  $\begin{array}{c}\frac{2250x}{x+24}=600\\ 2250x=600x+14400\\ 1650x=14400\\ x\approx 8.73\end{array}$

To get the new miles per gallon, add the value of xx which is 8.73 to the original miles per gallon which is 24.

  $8.73+24=32.73$

Therefore, the required miles is $32.73$ miles par gallons.